mfd: 88pm80x: Double shifting bug in suspend/resume
authorDan Carpenter <dan.carpenter@oracle.com>
Thu, 4 Aug 2016 05:26:56 +0000 (08:26 +0300)
committerGreg Kroah-Hartman <gregkh@linuxfoundation.org>
Sun, 16 Oct 2016 16:03:38 +0000 (18:03 +0200)
commit 9a6dc644512fd083400a96ac4a035ac154fe6b8d upstream.

set_bit() and clear_bit() take the bit number so this code is really
doing "1 << (1 << irq)" which is a double shift bug.  It's done
consistently so it won't cause a problem unless "irq" is more than 4.

Fixes: 70c6cce04066 ('mfd: Support 88pm80x in 80x driver')
Signed-off-by: Dan Carpenter <dan.carpenter@oracle.com>
Signed-off-by: Lee Jones <lee.jones@linaro.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@linuxfoundation.org>
include/linux/mfd/88pm80x.h

index d409ceb2231ec1908842416a9de1ddb62bef2709..c118a7ec94d6f15b25f558dbd171336da400a9f6 100644 (file)
@@ -350,7 +350,7 @@ static inline int pm80x_dev_suspend(struct device *dev)
        int irq = platform_get_irq(pdev, 0);
 
        if (device_may_wakeup(dev))
-               set_bit((1 << irq), &chip->wu_flag);
+               set_bit(irq, &chip->wu_flag);
 
        return 0;
 }
@@ -362,7 +362,7 @@ static inline int pm80x_dev_resume(struct device *dev)
        int irq = platform_get_irq(pdev, 0);
 
        if (device_may_wakeup(dev))
-               clear_bit((1 << irq), &chip->wu_flag);
+               clear_bit(irq, &chip->wu_flag);
 
        return 0;
 }
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